Question 1142000
.
<pre>

You need to maximize the objective function P(x,y) = x + 6y under given restrictions.


The feasible domain is shown below.


It is  a quadrilateral in the first quadrant  (x >= 0,  y >= 0)  restricted 
by the red line  x + y = 4  and the green line  2x + y = 7.



{{{graph( 330, 330, -2, 10, -2, 10,
          4-x,  7-2x
)}}}


Plots  x + y = 4  (red) and  2x + y = 7 (green)



The vertices of this quadrilateral are

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(X0,Y0) = (0,0)     (the origin of the coordinate system);

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(X1,Y1) = (0,4)     (red line Y-intercept);

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(X2,Y2) = (3,1)     (intersection point of the straight lines y = 4-x and  y = 7-2x );

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(X3,Y3) = (3.5,0)   (green line X-intercept)


Calculate the objective function at these points

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;P(X0,Y0) = 0 + 6*0   = 0;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;P(X1,Y1) = 0 + 6*4   = 24;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;P(X2,Y2) = 3 + 6*1   =  9;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;P(X3,Y3) = 3.5 + 6*0 =  3.5.


Select one of these points where the objective function is maximal. In our case this point is (X1,Y1) = (0,4).


This point gives your optimal solution x = 0,  y = 4.


The maximum objective function value is 24.
</pre>

Solved.


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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-minimax-problems-by--the-Linear-Programming-method.lesson>Solving minimax problems by the Linear Programming method</A> 

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